Codeforces Round #262 (Div. 2)

Codeforces Round #262 (Div. 2)

A：水题，直接不断模拟即可

B：由于s(x)大小最大到1e9,所以数位和最多为81，这样只要枚举s(x)，就只要枚举1到81即可，然后在计算出x，判断是否符合，符合就加进答案

C：二分高度，然后判断的时候for一遍，每次不符合的位置就去浇水，从左往右推一遍即可

D：构造，如果k >= 5, 那么就可以直接放偶数，奇数，偶数，奇数，由于偶数和偶数+1异或必然为1，所以这样放4个异或和就到最小的0了，然后k = 1，2，4都可以特判到，关键在于k = 3的情况下如何去构造，其实只要找到l的最高位，记下这个数字为p，如果3
p不超过r就可以构造了，可以选l, l + p, 3 p构造出来就是0，如果不行，那么肯定选2个异或和为1是最优

代码：

A:

#include <cstdio>
#include <cstring>
#include <cstdlib>

int n, m;

int main() {
int ans = 0;
scanf("%d%d", &n, &m);
while (n >= m) {
ans += n - n % m;
int num = n / m;
n = n % m + num;
}
printf("%d\n", ans + n);
return 0;
}

B:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <iostream>
using namespace std;

typedef long long ll;

ll a, b, c;
set<ll> ans;
ll out[105], on = 0;

ll pow1(ll x, ll k) {
ll ans = 1;
for (ll i = 0; i < k; i++)
ans *= x;
return ans;
}

bool judge(ll num, ll x) {
ll sum = 0;
while (num) {
sum += num % 10;
num /= 10;
}
return sum == x;
}

int main() {
scanf("%lld%lld%lld", &a, &b, &c);
for (ll i = 0; i <= 81; i++) {
ll tmp = b * pow1(i, a) + c;
if (tmp > 0LL && tmp < 1000000000LL && judge(tmp, i))
ans.insert(tmp);
}
for (set<ll>::iterator it = ans.begin(); it != ans.end(); it++)
out[on++] = *it;
printf("%lld\n", on);
for (ll i = 0; i < on - 1; i++)
printf("%lld ", out[i]);
if (on)
printf("%lld\n", out[on - 1]);
return 0;
}

C:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 100005;
const int INF = 1100000000;

int n, m, w;
int a
, tmp
;

bool judge(int h) {
int cnt = 0;
memset(tmp, 0, sizeof(tmp));
for (int i = 0; i < n; i++) {
if (i) tmp[i] += tmp[i - 1];
if (a[i] + tmp[i] < h) {
int add = h - tmp[i] - a[i];
cnt += add;
tmp[i] += add;
tmp[min(i + w, n)] -= add;
}
if (cnt > m) return false;
}
return true;
}

int solve() {
int l = 0, r = INF;
while (l < r) {
int mid = (l + r) / 2;
if (judge(mid)) l = mid + 1;
else r = mid;
}
return l - 1;
}

int main() {
scanf("%d%d%d", &n, &m, &w);
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
printf("%d\n", solve());
return 0;
}

D:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;

ll l, r, ans[10];
int k, an = 0;

void build() {
if (k == 1) ans[an++] = l;
else if (k == 2) {
if (r - l + 1 == 2) {
if (l < (l^r)) ans[an++] = l;
else {ans[an++] = l; ans[an++] = r;}
}
else {
if (l%2) l++;
ans[an++] = l; ans[an++] = l + 1;
}
}
else if (k == 3) {
ll p = 1;
while (p <= l) p *= 2;
p /= 2;
if (3 * p <= r) {
ans[an++] = l;
ans[an++] = l + p;
ans[an++] = 3 * p;
}
else {
if (l%2) l++;
ans[an++] = l;
ans[an++] = l + 1;
}
}
else if (k == 4) {
if (r - l + 1 > 4) {
if (l % 2) l++;
for (an = 0; an < 4; an++)
ans[an] = l + an;
}
else {
ll Min = l;
ans[an++] = l;
for (int i = 1; i < 16; i++) {
ll sum = 0;
for (int j = 0; j < 4; j++) {
if (i&(1<<j)) sum ^= (l + j);
}
if (sum < Min) {
Min = sum;
an = 0;
for (int j = 0; j < 4; j++) {
if (i&(1<<j))
ans[an++] = (l + j);
}
}
}
}
}
else {
if (l % 2) l++;
for (an = 0; an < 4; an++)
ans[an] = l + an;
}
}

int main() {
scanf("%lld%lld%d", &l, &r, &k);
build();
ll sum = 0;
for (int i = 0; i < an; i++)
sum ^= ans[i];
printf("%lld\n", sum);
printf("%d\n", an);
for (int i = 0; i < an - 1; i++)
printf("%lld ", ans[i]);
printf("%lld\n", ans[an - 1]);
return 0;
}